I want to understand why the following function has a weak derivative in two or three dimensions: $w(x) = \ln |\ln|x|| , x \in B_{1/2}(0)$. Can I say that if I have a strong derivative (except for the point 0), then it is shown? If yes, why is the dimension important? I computed the following derivative $\nabla w(x) = \frac{x}{(\ln|x|) |x|^2}$. Is it right?

Thanks a lot!

The definition of a weak derivative is: Let $w \in L^1_{\text{loc}}$. Then $u \in L^1_{\text{loc}}(\Omega)$ is called partial weak derivative if $\forall \phi \in D(\Omega)$:

$\int_\Omega w \partial_i \phi = -\int_\Omega \partial_i u \phi $. If this holds for all partial derivatives, $u$ is called weak derivative.

  • $\begingroup$ Did you verified the conditions $w,\nabla w\in L^1_{loc}(B_{1/2}(0))$? $\endgroup$ – Tomás Jan 2 '14 at 13:43
  • $\begingroup$ You can use spherial coordinates. $\endgroup$ – Tomás Jan 2 '14 at 13:59
  • $\begingroup$ Yes, your derivative is right (if I do not calculate it wrong). Before answer your last questions, I will ask your another questions: can you please write in your post the definition of weak derivative? $\endgroup$ – Tomás Jan 2 '14 at 14:11
  • $\begingroup$ Nice. Now, answering your last question, the only thing that remains to verify is the integral identity, but because the set where $w$ is not differentiable is a set with zero measure, then, the integral identity is true. $\endgroup$ – Tomás Jan 2 '14 at 14:56
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    $\begingroup$ The definition should say $\int_\Omega w \partial_i \phi = -\int_\Omega u \phi$, because $u$ is already thought of as a derivative. $\endgroup$ – Post No Bulls Jan 4 '14 at 16:41

The double log has a weak derivative in all dimensions $\ge 2$. The restriction to $n=2,3$ is not necessary.

Can I say that if I have a strong derivative (except for the point 0), then it is shown?

Depends on what you mean by strong derivative (there is such a term for differentiation of $L^p$ functions); I think you meant the pointwise derivatives, in which case the answer is no. Having a pointwise derivative, even at every point, is not enough to conclude that the weak derivative exists.

I will outline a different approach: write $w$ as an $L^1$ limit of functions $w_n$ for which the $x_i$ partial derivative, denoted $u_{i,n}$, exists, and converges in $L^1$ as $n\to\infty$. Then you can pass to the limit on both sides in $$\int_\Omega w_n \partial_i \phi = -\int_\Omega u_{i,n} \phi$$ A natural choice for $w_n$ is truncation $w_n=\min(w,n)$. To use it, you need to know that Lipschitz functions have weak derivatives. But the convergence part is straightforward.

Otherwise, you can use $w_n=\eta_n w$ where $\eta_n(x)=\eta(2^n x)$ and $\eta$ is a smooth function on $\mathbb R^d$ that vanishes when $|x|<1/2$ and is $1$ for $|x|>1$. Then there are longer estimates for establishing convergence of derivatives in $L^1$.


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